Physics - Electrostatics Question with Solution | TestHub

Here,${\text{C}}_1=\frac{{\epsilon }_0{\text{K}}_1\text{A}}{2\text{d}}+\frac{{\epsilon }_0{\text{K}}_2\text{A}}{2\text{d}}$

$=\frac{{\epsilon }_0\text{KA}}{2\text{d}}+\frac{{\epsilon }_0\text{KA}}{8\text{d}}$$\left(\begin{array}{cc}\text{Given :}& {\text{K}}_1=\text{K}\\ \text{and}& {\text{K}}_2=\frac{\text{K}}{4}\end{array}\right)$

$=\frac{{\epsilon }_0\text{KA}}{\text{d}}\left(\frac{1}{2}+\frac{1}{8}\right)=\frac{{\epsilon }_0\text{KA}}{\text{d}}\left(\frac{4+1}{8}\right)$

$=\frac{{5\epsilon }_0\text{KA}}{8\text{d}}$...... (i)

and${\text{C}}_2=\frac{{\epsilon }_0\text{A}}{\frac{\text{d}}{2{\text{K}}_1}+\frac{\text{d}}{2{\text{K}}_2}}=\frac{{\epsilon }_0\text{A}}{\frac{\text{d}}{2\text{K}}+\frac{\text{d}}{\frac{2\text{K}}{4}}}$

$=\frac{\left(2\text{K}\right){\epsilon }_0\text{A}}{5\text{d}}$.... (ii)

Dividing eqs. (i) by (ii)

$\frac{{\text{C}}_1}{{\text{C}}_2}=\frac{\frac{5{\epsilon }_0\text{KA}}{8\text{d}}}{\frac{2\text{K}{\epsilon }_0\text{A}}{5\text{d}}}=\frac{5\times 5}{2\times 8}=\frac{25}{16}$